tma4110 - calculus 3 - lecture 3

TMA4110 - Calculus 3Lecture 3

Toke Meier CarlsenNorwegian University of Science and Technology

Fall 2012

www.ntnu.no TMA4110 - Calculus 3, Lecture 3

Review of last week

Last week weintroduced complex numbers, both in a geometric way and inan algebraic way,defined Re(z), Im(z), |z| and arg(z) for a complex number z,defined addition and multiplication of complex numbers,defined complex conjugation,introduced polar representation of complex numbers,computed powers of complex numbers,defined and computed roots of complex numbers,solved complex, second degree equations.

www.ntnu.no TMA4110 - Calculus 3, Lecture 3, page 2

Solutions to second degree equations

If a,b, c are complex numbers and a2 6= 0, then the solutions to the

equation az2 + bz + c = 0 are z =−b ±

√b2 − 4ac

2a.


Problem 1 from the exam from June 2012

Solve w2 = (−1 + i√

3)/2. Find all solutions of the equationz4 + z2 + 1 = 0 and draw them in the complex plane. Write thesolutions in the form x + iy .


The fundamental theorem of algebra

Every complex polynomial of degree 1 or higher has a least onecomplex root.


Roots of real polynomials

If w is a root of a real polynomial∑n

k=0 akzk , then w is also a rootof∑n

k=0 akzk .


Exercise 34 on page xxvii

Check that z1 = 1−√

3i is azero of P(z) = z4−4z3+12z2−16z+16, and find all the zeros of P.


Complex functions

A complex function f is a rule that assigns a unique complexnumber f (z) to each number z in some set of complex numbers(called the domain of f ).

Examples of complex functionsf (z) = Re(z)g(z) = Im(z)h(z) = |z|j(z) = Arg(z)k(z) = zp(z) = z2 − 4z + 6


Graphic representations of complexfunctions

We cannot draw the graph of a complex function since wewould need 4 dimensions to do that.Instead, we can graphically represent the behavior of acomplex function w = f (z) by drawing the z-plane and thew-plane separately, and showing the image in the w-plane ofcertain, appropriately chosen set of points in the z-plane.


Example

Consider the function f (z) = −2iz.Arg(−2i) = −π/2 and | − 2i | = 2, so f maps the region1/2 ≤ |z| ≤ 1, 0 ≤ arg(z) ≤ π/2 to the region 1 ≤ |z| ≤ 2,−π/2 ≤ arg(z) ≤ 0.


Example

Consider the function g(z) = z2.g maps the region 0 ≤ |z| ≤ 1/2, π/2 ≤ arg(z) ≤ π to the region0 ≤ |z| ≤ 1/4, π ≤ arg(z) ≤ 2π.


Complex functions

Limits, continuity and differentiability of complex functions can bedefined just as for real functions.

Examples of complex functionsEvery complex polynomial is differentiable, and hencecontinuous.The functions f (z) = Re(z), g(z) = Im(z), h(z) = |z| andk(z) = z are continuous, but not differentiable.The function j(z) = Arg(z) is not continuous.


The exponential function

One can show that the series∞∑

n=0

zn

n!converges absolutely for all

complex numbers z.

We denote the sum of∞∑

n=0

zn

n!as the exponential function ez .

ez is also the limit limn→∞

(1 +

zn

)n.


The exponential function and cos and sin

If y is a real number, then

eiy =∞∑

n=0

(iy)n

n!=∞∑

n=0

(iy)2n

(2n)!+∞∑

n=0

(iy)2n+1

(2n + 1)!=

∞∑n=0

(−1)ny2n

(2n)!+ i

∞∑n=0

(−1)ny2n+1

(2n + 1)!= cos(y) + i sin(y).



1

y

eiy = cos(y) + i sin(y)



One can show that ez1+z2 = ez1ez2 . It follows thatez = exeiy = ex(cos y + i sin y) for z = x + iy .



ex

ez = ex(cos(y) + i sin(y))

yex


Example

ez maps the region 0 ≤ Re(z) ≤ 1/2, 0 ≤ Im(z) ≤ π/4 to theregion 0 ≤ |z| ≤ e1/2, 0 ≤ arg(z) ≤ π/4.


The exponential function and polarrepresentation

If z 6= 0, then

z = |z|(cos(arg(z))+ i sin(arg(z))) = eln(|z|)ei arg(z) = eln(|z|)+i arg(z).

The exponential function is not injective (becauseex+iy = ex+i(y+2π)), and does therefore not have an inverse.


Properties of the exponential function

If z = x + iy , thenez = ez

Re(ez) = ex cos yIm(ez) = ex sin y|ez | = ex

arg(ez) = yOne can also show that ez is differentiable and that d

dz ez = ez .


Sine and cosine


n=0

(−1)nz2n

(2n)!and

∞∑n=0

(−1)nz2n+1

(2n + 1)!converge absolutely for all complex numbers z.


n=0

(−1)nz2n

(2n)!as cos(z), and the sum of

∞∑n=0

(−1)nz2n+1

(2n + 1)!as sin(z).


Properties of sin and cos

If z is a complex number, then

cos z =eiz + e−iz

2and sin z =

eiz − e−iz

2i.

sin and cos are periodic with period 2π.sin and cos are differentiable and d

dz sin z = cos z andddz cos z = − sin z.


Hyperbolic sine and cosine


n=0

z2n

(2n)!and

∞∑n=0

z2n+1

(2n + 1)!converge

absolutely for all complex numbers z.


n=0

z2n

(2n)!as cosh(z), and the sum of

∞∑n=0

z2n+1

(2n + 1)!as sinh(z).


Properties of sinh and cosh

If z is a complex number, then

cosh z =ez + e−z

2and sinh z =

ez − e−z

2.

sinh and cosh are periodic with period 2πi .sinh and cosh are differentiable and d

dz sinh z = cosh z andddz cosh z = sinh z.


Problem 1 from the exam from August2011

Find all complex numbers z such that Im(−z + i) = (z + i)2. Drawthe solutions on a diagram.


Problem 1 from the exam from August2012

Write all of the solutions of z3 = 1 in the form z = x + iy . Write thesolutions of z3 = −3+i√

2(2+i)in the form z = x + iy and draw the

solutions in the complex plane.


tma4110 - calculus 3 - lecture 3

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